2t^2-16t+40=10

Solution for 2t^2-16t+40=10 equation:

2t^2-16t+40=10

We move all terms to the left:

2t^2-16t+40-(10)=0

We add all the numbers together, and all the variables

2t^2-16t+30=0

a = 2; b = -16; c = +30;
Δ = b2-4ac
Δ = -162-4·2·30
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*2}=\frac{12}{4}
=3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*2}=\frac{20}{4}
=5
$